Easy | General |

In a dark room, there is a deck of 52 cards, with exactly 10 cards facing up, rest facing down. You need to split this into two decks with equal number of cards facing up!

Hint

The question can be solved even when up facing cards are 11. The split decks need not have equal number of up-facing cards.

Solution

Create a deck of k cards randomly from the original 52-card deck, (k=10 here) and then turn over the k-card deck. Goal is achieved!

Source: Quant Interview

Enable Like and Comment Easy | Strategy |

A certain town comprises of 100 married couples. Some husbands secretly cheat on their wives. All wives know about the nature of every husband except their own. When a wife concludes that her husband cheated, she kicks her husband into the street at midnight. All husbands remain silent about their secret. One day, the mayor of the town announces to the whole town that there is at least 1 cheating husband in the town. After announcement, no one talks, waiting for someone to get kicked. Till 9th night from announcement, no husband was kicked, but on the 10th night, some husbands got kicked out simultaneously. How many are they?

Hint

What happens if only one husband cheated?

Solution

It must be 10 husbands kicked out.

If there was only 1 cheating husband in the town, there will be 99 women who know exactly who the cheater is. The 1 remaining woman, who is being cheated on, would have assumed there are no cheaters. But now that the mayor has confirmed that there is at least one cheater, she realizes that her own husband must be cheating on her. So her husband gets kicked on the day of the announcement.

Now let’s assume there are 2 cheaters in the town. There will be 98 women in the town who know who the 2 cheaters are. The 2 wives, who are being cheated on, would think that there is only 1 cheater in the town. Since neither of these 2 women know that their husbands are cheaters, they both do not report their husbands in on the day of the announcement. The next day, when the 2 women see that no husband was kicked, they realize that there could only be one explanation – both their husbands are cheaters. Thus, on the second day, 2 husbands are kicked.

Through induction, it can be proved that when this logic is applied to n cheating husbands, they are all kicked on the n th day after the mayor’s announcement. Hence it must be 10 husbands kicked in our case.

If there was only 1 cheating husband in the town, there will be 99 women who know exactly who the cheater is. The 1 remaining woman, who is being cheated on, would have assumed there are no cheaters. But now that the mayor has confirmed that there is at least one cheater, she realizes that her own husband must be cheating on her. So her husband gets kicked on the day of the announcement.

Now let’s assume there are 2 cheaters in the town. There will be 98 women in the town who know who the 2 cheaters are. The 2 wives, who are being cheated on, would think that there is only 1 cheater in the town. Since neither of these 2 women know that their husbands are cheaters, they both do not report their husbands in on the day of the announcement. The next day, when the 2 women see that no husband was kicked, they realize that there could only be one explanation – both their husbands are cheaters. Thus, on the second day, 2 husbands are kicked.

Through induction, it can be proved that when this logic is applied to n cheating husbands, they are all kicked on the n th day after the mayor’s announcement. Hence it must be 10 husbands kicked in our case.

Source: Common

Enable Like and Comment Easy | Probability |

A father claims about snowfall last night. First daughter tells that the probability of snowfall on a particular night is 1/8. Second daughter tells that 5 out of 6 times the father is lying! What is the probability that there actually was a snowfall?

Hint

Conditional Probability or Baye's Theorem

Solution

Let S = Snowfall occurred, and C = Claim

Probability of (Snowfall given Claim) = P(S | C) = P(C|S)*P(S)/P(C)

Now, P(C|S) = 1/6, P(S) = 1/8

P(C ) = P(true claim) + P(False Claim) = P(C|S)*P(S) + P( false claim|no snow)*P(no snow)

This is same as [1/6*1/8]/[ 1/6*1/8 + 7/8*5/6] = 1/36

Probability of (Snowfall given Claim) = P(S | C) = P(C|S)*P(S)/P(C)

Now, P(C|S) = 1/6, P(S) = 1/8

P(C ) = P(true claim) + P(False Claim) = P(C|S)*P(S) + P( false claim|no snow)*P(no snow)

This is same as [1/6*1/8]/[ 1/6*1/8 + 7/8*5/6] = 1/36

Source: Written Test

Enable Like and Comment Easy | General |

You are in a game against devil, on a perfectly round table and with an infinite pile of pennies. He says, "OK, we'll take turns putting one penny down, no overlapping allowed, and the pennies must rest flat on the table surface. The first guy who can't put a penny down loses." You can go first. How will you guarantee victory?

Hint

Try the problem if coins have unit radii, and table has radius 1 & 3.

Solution

You place a penny right in the center of the table. After that, whenever the devil places a penny on the table, mimic his placement by placing a penny diametrically opposite and at same distance from center. If he has a place to place a penny, so will you. The devil will run out of places to put a quarter before you do.

Easy | Probability |

Two witches make a nightly visit to an all-night coffee shop. Each arrives at a random time between 0:00 and 1:00. Each one of them stays for exactly 30 minutes. On any one given night, what is the probability that the witches will meet at the coffee shop?

Hint

Graph Shading

Answer

3/4

Solution

Plot the graph of X and Y, where X and Y denote the time each witch arrives. This will form a 60x60 square for total feasible region. The probability that they meet is |X-Y|<= 30. Draw this as favorable region, by joining line (30,0) with (60,30) and (0,30) with (30,60). Clearly the interior of this region has area 3/4 th of total. Hence the probability = 3/4 = 0.75

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